ĐK:\(\left[{}\begin{matrix}x\ge\sqrt{10}\\x\le-\sqrt{10}\end{matrix}\right.\)
PT\(\Leftrightarrow x^2-10+2\sqrt{x^2-10}-3=0\)
Đặt \(t=\sqrt{x^2-10}\left(t\ge0\right)\)
\(\Leftrightarrow t^2+2t-3=0\)
Theo ĐK: \(t=1\Leftrightarrow\sqrt{x^2-10}=1\)
\(\Leftrightarrow x^2=11\Rightarrow x=\pm\sqrt{11}\) (tm)
Bài 3: Giải phương trình
2) x2 - 2\(\sqrt{13}x\) + 13= 0
3) (x+2)\(\sqrt{x-3}=0\)
4) x+ \(\sqrt{x-2}=2\sqrt{x-1}\)
5) \(\sqrt{9-12x+4x^{2^{ }}}=4\)
1) x2 -5 =0
Rút gọn các biểu thức sau:
a) \(\dfrac{4}{\sqrt{11}-3}-\dfrac{5}{4+\sqrt{11}}\)
b) \(\left(\dfrac{3\sqrt{x}}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{x+6\sqrt{x}+9}\) với x>0;x\(\ne\)4
giải hpt \(\left\{{}\begin{matrix}x^2-\left(x-1\right)\sqrt{y+2}+3x=4\\\sqrt{x^2+8x+13}+\sqrt{10-y}=3\end{matrix}\right.\)
Giải phương trình bằng phương pháp bất đẳng thức
1, \(\sqrt{x^2-6x+11}+\sqrt{x^2-6x+13}+\sqrt[4]{x^2-4x+5}=3+\sqrt{2}\)
2, \(\sqrt{x-10}+\sqrt{30-x}=x^2-40x+400+2\sqrt{10}\)
3, \(x^2-3x+3,5=\sqrt{\left(x^2-2x+2\right)\left(x^2-4x+5\right)}\)
4, \(\sqrt{5x^3+3x^2+3x-2}=\dfrac{x^2}{2}+3x-\dfrac{1}{2}\)
5, \(2\sqrt{7x^3-11x^2+25x-12}=x^2+6x-1\)
Tìm GTLN của:
\(A=13\sqrt{x^2-x^4}+9\sqrt{x^2+x^4}\) với \(0\le x\le1\)
a)\(3x^2-13+37=8\left(x-3\sqrt{x+2}\right)\)
b)\(3x^2+3x+18=10\sqrt{x^3+8}\)
Giải hệ phương trình:
1, \(\left\{{}\begin{matrix}\left(17-3x\right)\sqrt{5-x}+\left(3y-14\right)\sqrt{4-y}=0\\2\sqrt{2x+y+5}+3\sqrt{3x+2y+11}=x^2+6x+13\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x\left(x+y\right)+\sqrt{x+y}=\sqrt{2y}\left(\sqrt{2y^3}+1\right)\\x^2y-5x^2+7\left(x+y\right)-4=6\sqrt[3]{xy-x+1}\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt[4]{32-x}-y^2+3=0\\\sqrt[4]{x}+\sqrt{32-x}+6y-24=0\end{matrix}\right.\)
Gpt: a) \(\sqrt[4]{3\left(x+5\right)}-\sqrt[4]{11-x}=\sqrt[4]{13+x}-\sqrt[4]{3\left(3-x\right)}\)
b) \(\frac{1+2\sqrt{x}-x\sqrt{x}}{3-x-\sqrt{2-x}}=2\left(\frac{1+x\sqrt{x}}{1+x}\right)\) c) \(\sqrt{x+1}+\frac{4\left(\sqrt{x+1}+\sqrt{x-2}\right)}{3\left(\sqrt{x-2}+1\right)^2}=3\)
d) \(\sqrt{\frac{x-2}{x+1}}+\frac{x+2}{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}=1\) e) \(2x+1+x\sqrt{x^2+2}+\left(x+1\right)\sqrt{x^2+2x+2}=0\)
f) \(\sqrt{2x+3}\cdot\sqrt[3]{x+5}=x^2+x-6\)
Rút gọn A = \(\left(\frac{x+2\sqrt{x}+4}{x\sqrt{x}-8}+\frac{x+2\sqrt{x}+1}{x-1}\right) :\left(3+\frac{1}{\sqrt{x}-2}+\frac{2}{\sqrt{x}+1}\right)\)
a, Rút gọn A b , Tìm x thỏa mãn A > 1 c,Tính A với \(x=\frac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt{2}}\)\(A=\frac{\sqrt{x}+1}{3\left(\sqrt{x}-1\right)}\)
